Confusion with dimension of cartesian vs tensor product I learned that given a vector space $V$, you can make a dual vector space $V^*$, which is the space of the linear maps from $V$ to $\mathbb{R}$. Both spaces have the same dimension D.
If we define the Cartesian product $V \times V$, then we can define the tensor space $V^* \times V^*$, which is the space of the linear maps from $V \times V$ to $\mathbb{R}$.
Question: Is $V^* \times V^*$ the dual of $V \times V$? It seems it is not, because the dimension of $V^* \times V^*$ is $D^2$, but the dimension of $V \times V$ is 2D. So, why aren't they duals? or are they and I said something wrong?
Note: One comments mentions that the issue is related to bilinear vs linear map. I am not sure to understand why or how, so if that is the case please be give me a little more of detail. Thanks.
Note 2:
This professor defined $V^* \times V^*$ as $(V \times V)^*$, that is, the dual space of the cartesian product of V with itself. By dual I mean the set of linear transformations from $V \times V$ to the reals.
Now, it clear that $V \times V$ has dimension 2D, where D is the dimension of V. What is the dimension of the above defined $V^* \times V^*$? He said one base is the set of linear transformations satisfying $(e_i, e_j)=\delta_{ij}$, but this base has dimension $D^2$, right? He also uses the dual to define tensors, which also makes me think the dimension is $D^2$. So I am still confused. Thanks.
 A: When we talk about a vector space $V$, we talk about both the set itself and its operations.

Question: Is $V^∗ \times V^∗$ the dual of $V \times V$?

The set $V^* \times V^*%$ contains all elements in the shape of $(f, f')$ where $f, f' \in V^*$. Since you only need to determine where the basis maps to in order to determine a linear map, we have the basis
$\{f_i\}$ for $V^*$ where $f_i$ maps the $i$st basis of $V$, $v_i$ to $1$ and maps other $v_j$ to $0$. The basis for $V^*$ has a length $D$.
Hence we have the basis $\{(f_i,\mathbf{0})\} \cup \{(\mathbf{0}, f_i)\}$ for $V^* \times V^*$ as a vector space. The basis has a length $2D$.
The dual of $V \times V$ contains all the linear map from $V \times V$ to its field $\mathbb{F}$, which has the basis $\{g_i\} \cup \{g'_i\}$, each $g_i$ maps $(v_i,\mathbf{0})$ to $1$ and each $g'_i$ maps $(\mathbf{0}, v_i)$ to $1$. The basis also has a length $2D$.
Note that the elements in $V^* \times V^*$ are a pair of linear maps, while the dual of $V \times V$ is a linear map that maps a pair of vectors in $V$. So they appear not to be the same at first, but you can treat them the same way.
If you are asking for $V^* \otimes V^*$, they do have dimension of $D^2$, and so does the dual of $V \otimes V$.
A: With the comments aiding to focus, the statement what
you are looking is the dual of $V\oplus V$ is $V^*\oplus V^*$, that is
$$(V\oplus V)^*\cong V^*\oplus V^*$$
This, also helps us not to involve with the tensor product.
